Yes, sorry i do understand that. But I have no idea how it could be answered with the information provided . When it states range I would assume that it would refer to the span of the data for the upper and lower control limits. However this just states that the ‘average’ is 95.75 with a sample of 20… so i’m completely unsure how you would go about workout out the range just from these figures.

If there is a way to answer that question I will have to admit I can’t think of a way to do it. You are correct – the usual problem of this type is to give a min and a max, compute their average and then divide the result by d2 for 20 samples (d2 comes from tables of percentage points of the distribution of the relative range) For 20 samples that would be 3.735.

I would say that if taken at face value – 95.75 is the mean and 20 is the sample size then there is no way to estimate the standard deviation and even if you did so the statement/question “what would an average of 95.75 show in terms of the range…” doesn’t make any sense.

A real stretch – If we assume the question implies that the average range is 95.75 then the estimate of the standard deviation based on 20 samples would be 25.6 but that number isn’t close to any of the choices and it still begs the question – what does the phrase “show in terms of the range” mean.

My guess would be that at least one part of the question went missing in action when it was transcribed.

@MBBinWI Actually I started to post a comment with respect to the overall quality of the test and what it might suggest relative to the quality of the training received. I find this question particularly worrisome since it seems to be the second poorly presented question in the material the OP is using.